If $cot x - tan x = 2$ , the generalised solution is (here, $n$ is integet)

x = \frac{n π}{2} + \frac{π}{8}

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x = \frac{n π}{4} + \frac{π}{16}

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x = 2 n π + \frac{π}{2}

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x = n π + \frac{π}{4}

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Solution

The correct option is A $x = \frac{n π}{2} + \frac{π}{8}$
We have, $cot x - tan x = 2$
$\Rightarrow \frac{cos x}{sin x} - \frac{sin x}{cos x} = 2$
$\Rightarrow \frac{{cos}^{2} x - {sin}^{2} x}{sin x cos x} = 2$
$\Rightarrow cos 2 x = 2 sin x cos x$
$\Rightarrow cos 2 x = sin 2 x$
$\Rightarrow tan 2 x = 1 = tan \frac{π}{4}$
$\Rightarrow 2 x = n π + \frac{π}{4}, n \in Z$
$\Rightarrow x = \frac{n π}{2} + \frac{π}{8}, n \in Z$ .

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If cotx−tanx=2, the generalised solution is (here, n is integet)

The correct option is A x=nπ2+π8We have, cotx−tanx=2⇒cosxsinx−sinxcosx=2⇒cos2x−sin2xsinxcosx=2⇒cos2x=2sinxcosx⇒cos2x=sin2x⇒tan2x=1=tanπ4⇒2x=nπ+π4,n∈Z⇒x=nπ2+π8,n∈Z.

If $cot x - tan x = 2$ , the generalised solution is (here, $n$ is integet)